misc
Power equation
1/r^2
goes to infinity when close
Gauss Law (describe it)
difference between electric field and electric potential
Electric field (E) is a vector quantity that represents the force per unit charge exerted on a test charge at a given point in space.
Electric potential (V) is a scalar quantity that represents the electric potential energy per unit charge at a point, describing the work done in moving a test charge from a reference point to that location without acceleration.
what is Ohms law
what is V = IR
what and what specify any field
div and curl
coulombs law
1/(4 pi * ε0) * (q*Q)/r^2
the value of a line integral is independent of what
path
You have two metal balls some distance d apart out in space. One is charged
to a potential V (relative to infinity). The balls are then connected together by
a wire for a brief period, and the wire is removed. What happens next?
They repel. For one ball to be at V means that charge was added
to it (its a capacitor after all). When the two identical balls are placed in
electrical contact, charge should immediately evenly distribute between both of
them (they are identical and charge is free to move around in a conductor).
When the wire is removed, the charge is trapped there, putting a net charge on
both balls. The result is repulsion
farad
whats voltage? whats current?
give a general description
Voltage is the electrical potential difference between two points in a circuit, driving the flow of charge.
Current is the rate at which electric charge flows through a conductor, measured in amperes (A).
A conducting spherical shell carries total charge +Q. A point charge +q is placed at the center of the cavity.
What is the electric field inside the conducting material?
Inside conductor: E=0
Free charges rearrange so the conductor remains an equipotential.
div of a curl =
and why?
zero
Physically, taking the divergence of a curl would mean measuring how much a purely rotational field "spreads out," which doesn't happen since rotation doesn't create a net "source" or "sink."
Couloumbs law takes a charge distribution and gives back the ____
Line integration takes the field and gives the ____
electric field
electricl potential
Capacitor in series rule
1/Ceq = 1/c1 + 1/c2 + 1/c3 ...
You build a circuit made entirely out of resistors that is powered by a fixed
voltage V. Somebody comes along and doubles all the resistance values. What
happens to the voltages inside the circuit, assuming the supply voltage stays
the same? (hint: what does current conservation have to say here?)
First, when the circuit is operating, every voltage needs to satisfy two rules:
current needs to be conserved everywhere and the sums of voltages around
closed loops need to equal zero. If we double all the resistances, you’d still have
a valid solution if you simply kept all the voltages the same and scaled all the
currents by 1/2.
You can make this a little less abstract by considering all the currents flowing
into the circuit and how they split up at a junction. Consider the fact that the
whole circuit is made of resistors. Thus, as far as the battery is concerned, it
can be modeled as a single effective resistance. If I double every resistance in
the network, then the effective resistance goes up by 2. Thus the battery would
input 1/2 the original current to the circuit. When that current hits the first
junction in the circuit, the part upstream is the same as the original voltage (its
still tied to V). The only way the downstream voltages can then work out so
that the total current flowing out of the junction is 1/2 the original value given that each of those resistance is also doubled is if the voltages all maintain their
original values. This chain of logic could be re-aplied at every junction where
the current splits, giving the answer that all the voltages must stay the same.
Suppose you have a ball of uniform positive charge density ρ. Where should
you place a positive point charge so that it experiences the maximum repulsion
from the sphere (feel free to draw a picture or answer in words)
You place it just in contact with the surface of the sphere. You could
answer this either by noting this is where the field is the highest or by noting
this is the closest you can get to the charges in the problem such that all of them
are arranged on one side: move away from the edge of the ball and the force
goes down because of 1/r^2, move inside of it and the force goes down because
some of the charge is pushing you the other way.
For an infinitely long rod with uniform charge density λ, the electric field at a distance r from the rod is given by this expression.
E = λ / (2*pi * r * ε)
Enclosed Charge: The charge enclosed by the Gaussian surface is the linear charge density λ multiplied by the length L of the rod segment inside the Gaussian surface:
Qenc =λL
Electric Flux: The electric flux through the cylindrical surface is given by the integral of the electric field E\mathbf{E}E over the surface. Since the electric field is radial and has the same magnitude at all points on the cylindrical surface, the flux simplifies to:
ΦE=E(2πrL)
now E(2πrL) = λL / e0
can solve for E
Consider the electric field of a point charge at the origin.
Where is:
Div E = 0
Div E = 0 everywhere but the origin, as div E = rho/e0
Curl E = 0 everywhere
Suppose you have resistors all with the value R and capacitors all with the same value C. Draw a circuit where the RC time constant would be RC/2
a resistor and two capacitors in series or a capacitor
in series with two resistors in parallel
You connect a resistor R and capacitor C in series to a battery V. At what time is the current maximum?
Immediately at t=0
Current:
I(0)=V/R
After that:
I(t)=V R e^−t/RC
Two solid spheres of equal radius overlap. Both have uniform charge density +ρ.
What is Div E in the overlap region?
2ρ/e0
A charge Q sits at a height d/2 above a square of side length d. What is the
flux through the square (hint: what are cubes made of?)
the flux is Q/6ε0. If the charge were enclosed by a cube, then Gauss’s
law in integral form gives us the flux directly R dAˆn· ∫ E = Qenc/ε0. Since there is
only charge Q, the flux thru a cube would be Q/ε0. But given the placement of
the charge, each face of the cube would see the same flux (they are symmetric).
So the flux through a single face (i.e. the square) would be Q/ε06
Gauss’s Law provides a fundamental relationship between charge and electric field, but electric potential offers a different perspective. Explain why electric potential is often more useful than the electric field in solving electrostatic problems, and describe a scenario where solving for potential first simplifies finding the electric field.
electric potential is a scalar field, making it easier to work with than the vector nature of the electric field, and once potential is known, the electric field can be found by taking its gradient. A common scenario is finding the electric field of a ring of charge along its central axis, where using potential first avoids direct vector summation and instead allows differentiation to obtain the field.
What is the RC constant
Rc constant is the time required to charge the capacitor through the resistor from an initial voltage of 0 to 63% of its input voltage